# Mean Value Theorem follow on question

• Jul 21st 2009, 07:45 AM
takeonme79
Mean Value Theorem follow on question
Hi, please could someone help me with the following

f(x) = square root (x), x>0
Apply the Mean Value Theorem to f(x) on the interval [100,102] to prove that

111/11 < sqrt (102) < 101/10

I've proved that sqrt (102) < 101/10 by saying that there exists c such that the gradient at c = ( f(102) - f(100) ) / (102-100), then by saying that the gradient at c is less than the gradient at x=100

Annoyingly, if i try saying the gradient at c is greater than the gradient at x=102 then I end up with exactly the same inequality. How do i prove the left hand inequality, 111/11 < sqrt (102)?

Thanks.
• Jul 21st 2009, 09:10 AM
Random Variable
let $f(x) = \sqrt{x}$

$\sqrt{102} - \sqrt{100} = \sqrt{102} - 10 = \frac {1}{2 \sqrt{c}} (102-100) = \frac {1}{\sqrt{c}}$

where $c \in [100,102]$

$100 < c < 102$

$10 < \sqrt{c} < \sqrt{102} < 11$

$\frac {1}{11} < \frac {1}{\sqrt{c}} < \frac {1}{10}$

which implies $\frac {1}{11} < \sqrt{102} - 10 < \frac {1}{10}$

$\frac {1}{11} + 10 < \sqrt{102} < \frac {1}{10} + 10$

$\frac {111}{11} < \sqrt{102} < \frac {101}{10}$
• Jul 21st 2009, 09:24 AM
takeonme79
Thanks so much, I would never have thought of introducing root (102) <11.

So I can use my original method as well, by saying the gradient at c is greater than at x=121.

Thanks again!